Question pattern: Kinematics data analysis
Original question 1 of 2 in this pattern
Acceleration from motion-sensor data
A motion sensor records the velocity of a cart. Each velocity value has an uncertainty of ±0.10 m s⁻¹; time uncertainty is negligible.
a) Use the first and last readings to estimate the cart’s acceleration.
b) Estimate the absolute uncertainty in this acceleration using the worst-case uncertainty in the change in velocity.
c) Use the constant-acceleration model to predict the velocity at t = 2.50 s.
d) Evaluate whether the readings support a constant-acceleration model.
| Time / s | Velocity / m s⁻¹ |
|---|---|
| 0.00 | 1.2 |
| 0.50 | 2.0 |
| 1.00 | 3.1 |
| 1.50 | 3.9 |
| 2.00 | 4.9 |
Worked answer
Solution
- Step 1
Acceleration from the full interval:
a = (4.9 - 1.2) / (2.00 - 0) = 1.85 m s⁻². - Step 2
Worst-case uncertainty in
Δvis0.10 + 0.10 = 0.20 m s⁻¹, soδa = 0.20 / 2.00 = 0.10 m s⁻². - Step 3
Report the estimate as approximately
a = (1.9 ± 0.1) m s⁻². - Step 4
Using the unrounded model,
v(2.50) = 1.2 + 1.85(2.50) = 5.83 m s⁻¹, or about 5.8 m s⁻¹. - Step 5
The model predicts 2.13, 3.05 and 3.98 m s⁻¹ at the intermediate times. The observed differences are −0.13, +0.05 and −0.08 m s⁻¹.
- Step 6
These small residuals are comparable with the ±0.10 m s⁻¹ measurement uncertainty, so the data support an approximately constant acceleration; they do not prove it is exactly constant.
Final answer
Acceleration ≈ (1.9 ± 0.1) m s⁻²; predicted v at 2.50 s ≈ 5.8 m s⁻¹; the residuals are consistent with approximately constant acceleration within the stated uncertainty.
Common mistake
Treating every small deviation from the model line as evidence that acceleration cannot be constant.
Concepts tested
- motion sensor
- gradient
- acceleration
- measurement uncertainty
- model evaluation
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